Fuse/supply capacitor calculations

Hello,

The problem seems simple: I have a circuit on 24V DC I'm trying to protect with a 250mA SMD fuse(*) in the supply line. The circuit itself needs some

50mA at most, and several PCB tracks and components would constitute a fire hazard at substantially more than a few hundred mA, so a 250mA fuse would be ideal.

*: See

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The trouble is that the fuse often blows upon switching on the supply voltage, probably due to the inrush current into a 220uF supply capacitor behind the fuse.

The fuse has a typical I^2t value of 0.000084, and a cold series resistance of 4 ohms (including wiring etcetera), so one would expect a worst case inrush current of 24/4=6A at the moment of switching on; at a constant 6A, charge time is some 0.9 milliseconds -- and as the datasheet specifies a hold time of 1 millisecond at only 1.3A, it's not surprising that the fuse should blow.

The point is, however, that this 6A very rapidly diminishes as the voltage over the capacitor rises, following an e-power curve, so the actual I^2t load on the capacitor is lower. First the theory question: Is there an easy calculation providing a link between a diminishing inrush current and an I^2t value?

The practical question is of course how to solve this particular problem. There appear to be no slow fuses in 1206 SMD cases, so I guess I'll have to resort to using a bigger type of fuse. I could also use a smaller capacitor (down to some 22uF) -- but the inrush current would stay the same, only with a correspondingly shorter charge time. So I'd need really need to know (calculate?) what happens to the fuse in these cases. But perhaps someone can come up with another good suggestion?

And oh: I have done some experiments with SMD resistors, but these turn out to be amazingly resilient to excess power -- I had a 1.5 ohm 1206 resistor desoldering itself after a minute or so of carrying almost 1 ampere of current, with only a minor increase in resistance. And at 2 amperes, it glowed red for several seconds before failing -- but by then I guess the charred PCB could take over the role of conductor.

Thanks in advance,

Richard Rasker

--
http://www.linetec.nl/
Reply to
Richard Rasker
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At only 50 mA, maybe you could use a resistor _and_ a fuse, along with cutting the capacitor by a factor of ten, if you can actually stand that. Seems like a much smaller capacitor could also have a higher ESR, and so a lower peak current.

-- John

Reply to
John O'Flaherty

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About the theoretical part, it's an RC circuit, and the resistor (fuse) voltage is V * e^(-t/(RC)). (The original 24V times a factor that goes from 1 to zero as time goes from zero to infinity). The current is that divided by the resistance R. The I^2t should be integral ( (V/R*e^(-t/RC))^2 dt, t = 0 to infinity).

With V = 24, R = 4ohms, C = 220uF, this is int( (24V/4ohms * e^(-t/(4ohm*220uF)) ^2 dt, t = 0 to infinity), which my TI-89 says is .01584 amps squared * seconds. Or square amps? With 22 uF, it's .001584. With 220 uF and an additional 4 ohm resistor (total 8 ohms), you get .00792 for I^2T. With 22 uF and 8ohms total, you get 0.000792. The value scales linearly with capacitance, and non- linearly with resistance.

This ignores any change in resistance of the parts due to heating, which would depend on their physical description.

-- John

Reply to
John O'Flaherty

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Sorry to stutter. It scales linearly up with capacitance, and down with resistance, unless I've done something wrong.

-- John

Reply to
John O'Flaherty

So put the fuse after the cap.

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And I did something wrong - I forgot to multiply by t in the integral of I^2T. The result is independent of resistance, and depends only on the capacitance, which makes sense, since the capacitance charges as current times time.. The higher the R, the lower the current but the longer the time. The value of I^2T for 24 V and 220 uF is 7*10^-6. So, with the same capacitor but a larger series resistance, the fuse is able to dissipate the heat, but the energy passed through is the same. The only thing worse than being wrong is staying wrong. Sorry I can't help on the theoretical question!

-- John

Reply to
John O'Flaherty

Hi Richard,

I2t = 0.5*(Ipeak)^2*(36% decay time).

So faster charging reduces the charging time, but in the end increases I2t. I assume that you switch from a low output impedance source. Is there any room/space/budget to add a current limiter (that "consumes" just 100mV under normal use), or a dV/dt limitation?

Best regards,

Wim PA3DJS

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On the off chance that anyone is still following my string of self- replies, I have to correct myself yet another time. Integration across time is already multiplying by time, and my original calculation of I2T, and the formulas, were correct. The total I*t to charge a capacitor is fixed, but the I^2*t isn't. I apologize to everyone for my dithering.

-- John

Reply to
John O'Flaherty

How about a Polyswitch self-resetting thermal fuse?

Tyco Electronics makes some of these in a 1206 size. As one example, their NANOSMDC020F-2 part will sustain a current flow 200 mA indefinitely, will trip at 420 mA, and works at up to 24 VDC.

Other vendors of polymer self-resetting PTC overcurrent protectors may have equivalent or similar parts.

--
Dave Platt                                    AE6EO
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Dave Platt

...

[snip]

I especially chose a low ESR cap of a relatively high value because it's hooked up to an SMPS, but I think I can get away with 22uF.

I suspected out so much, although I hadn't started on calculations yet.

I'd say amps squared seconds. I think I'll do the math myself as well, if only to brush up on my math skills, but this 0.016 is indeed two hundred times higher than the specified I^2t value of the fuse. This, by the way, is at least an order of magnitude lower than the I^2t of regular glass fuses, even the fast ones (which also explains why I never ran into this problem before, having built hundreds of fuse-protected circuits ...)

Well, thanks a lot for your elaborate reply (and all the followups :-). I'll probably just take a slower type of fuse, perhaps combined with a smaller capacitor.

Thanks again, best regards,

Richard Rasker

--
http://www.linetec.nl/
Reply to
Richard Rasker

I have looked at those, but their popularity (and thus availability) seems to be waning -- all the types offered by Farnell, for instance, are no longer in production. But thanks for the suggestion anyway.

Richard Rasker

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http://www.linetec.nl/
Reply to
Richard Rasker

Thanks, but that won't do. The techical requirements for this circuit are that the fuse must be placed before anything else, e.g. to prevent trouble if the supply is hooked up the wrong way round.

Richard Rasker

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http://www.linetec.nl/
Reply to
Richard Rasker

You're welcome. I note that Wimpie's reply is exactly right (and simpler), with R*C as the 36% discharge time - it's exactly what that integral evaluates to. It can also be expressed as .5* V^2 * C/R, which shows the direct linear dependence on C and inverse on R.

-- John

Reply to
John O'Flaherty

[snip]

With the current's influence squared and the time factor linear, this means that an increased current is far worse than a longer charging time.

The 24 volts supply voltage is from a big ship's batteries, capable of delivering a few thousand amps, so yes, I'd think we can call that a low impedance source :-)

There is room for a current limiter, but I think the low-tech solution of simply using a slower fuse is better. I think I'll also run some tests with several types of fuses and several capacitor values, having a relay switch between charging and discharging the capacitor through the fuse once a second for an hour or so. This should give some more information on a practical combination of values. If anything interesting comes out of this, I'll post it here.

Thanks for your reply, best regards,

Richard Rasker

--
http://www.linetec.nl/
Reply to
Richard Rasker

Hi Richard,

Did you also take into account that when the switching action occurs frequently the expected life of your electrolytic capacitor may reduce significantly.

Depending on the inductance of the feed line and resistive losses (probably low), you may also get a voltage spike across the input electrolytic capacitor (the better the capacitor, the higher the spike). You might run a simplified (spice) simulation to see the effect for yourself.

Some months ago a client had a similar fuse problem. He also wanted to use an SMT fuse (space problem). I2t rating of the SMT fuses were far too low (so they use a glass fuse now).

Best regards,

Wim PA3DJS

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Reply to
Wimpie

The circuit is switched on and off a few times a week at most, so I don't think this will be a serious problem.

I couldn't fit in a glass fuse, but inserting a series resistor of a dozen ohms fixed it for me -- the resulting extra voltage drop is no problem, and even after multiple thousand on/off cycles (with a 0.5Hz oscillator and a relay clicking away for well over an hour) nothing failed.

Anyway, I've learned a lot about (SMT) fuses, and in particular that the specification "fast" doesn't say anything, as the I^2t may vary orders of magnitude among different types of "fast" fuses with the same current rating.

Thanks once again, best regards,

Richard Rasker

--
http://www.linetec.nl/
Reply to
Richard Rasker

I^2t is proportional to the capacitor's stored energy, 0.5QV^2, so this theory suggests that the only solution is to reduce the cap size. In reality, a fuse will pass infinite energy (given a long enough time), so as you have concluded, one solution is to stretch out the charge time somehow to get beyond the fuse I^2t region.

Is the device load constant enough that you could put a resistor ahead of the cap and load? Then, bump up the supply voltage to compensate for the IR drop. Or place a regulator downstream of the cap.

If you can't put up with the power loss through the resistor, there may be another way. How is this device powered up? Is it possible to bypass the fuse and power switch* with a high value resistance that will pre-charge the cap? The resistor can be selected such that a short to ground downstream of it will draw a low enough current so that protecting it with a fuse is not necessary.

*Could be a series MOSFET delayed to power up once the cap voltage has risen.
--
Paul Hovnanian	paul@hovnanian.com
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